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December 9th, 2016, 08:15 PM   #1
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canonical base problem

The attached figure illustrates a rectangular coordinate system xy generated by the unit vectors of the canonical base $\displaystyle \vec{i}$ and $\displaystyle \vec{j}$ and a system of coordinates also rectangular $\displaystyle {x}' {y}'$ generated by the unit vectors $\displaystyle {\vec{i}}'$ and $\displaystyle {\vec{j}}'$ obtained of the rotation of the rectangular coordinate system xy counterclockwise by an angle $\displaystyle \theta$ = $\displaystyle \frac{\Pi}{6}$. Find the coordinates $\displaystyle {x}' {y}'$ for points whose coordinates x y are (3, -10). ( Use $\displaystyle \cos (\frac{\Pi}{6})$ = $\displaystyle \frac{sqrt{3}}{2}$ and $\displaystyle \sin (\frac{\Pi}{6})$ = $\displaystyle \frac{1}{2}$ Recall that $\displaystyle x\vec{i} + y\vec{j}$ = $\displaystyle {x}'{\vec{i}}' + {y}'{\vec{j}}'$ )
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 December 10th, 2016, 05:50 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 667 Draw lines from (3, -10) perpendicular to the new coordinate axes which are the lines [tex]y= \frac{\sqrt{3}}{3} x [tex] and $y= -\sqrt{3}x$. What are the distances from (3, -10) to those two lines? Do you see how those give the x' and y' coordinates?
December 10th, 2016, 10:22 AM   #3
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Quote:
 Originally Posted by Country Boy Draw lines from (3, -10) perpendicular to the new coordinate axes which are the lines [tex]y= \frac{\sqrt{3}}{3} x [tex] and $y= -\sqrt{3}x$. What are the distances from (3, -10) to those two lines? Do you see how those give the x' and y' coordinates?
I can not understand how to do it.

 December 10th, 2016, 10:58 AM #4 Senior Member     Joined: Sep 2015 From: CA Posts: 1,300 Thanks: 664 The usual way to do these changes of basis is to project points in the first coordinate system onto the new basis vectors. Here our first set of basis vectors is $v_1 = \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)$ $v_2 = \left(\dfrac{-1}{2},\dfrac{\sqrt{3}}{2}\right)$ and our second set of basis vectors is $v_1^\prime = (1,0)$ $v_2^\prime = (0,1)$ The point $p=(p_x, p_y)$ in the first coordinate system will have coordinates $(p_x v_1 \cdot v_1^ \prime + p_y v_2 \cdot v_1^\prime, ~p_x v_1 \cdot v_2^\prime + p_y v_2 \cdot v_2^\prime)$ in the second coordinate system. and note this can be rewritten as $\begin{pmatrix} v_1 \cdot v_1^\prime &v_2\cdot v_1^\prime \\ v_1 \cdot v_2^\prime &v_2 \cdot v_2^\prime \end{pmatrix} \begin{pmatrix}p_x \\ p_y\end{pmatrix} = R p$ $R$ is referred to as a rotation matrix. So compute $R$ and find $R \begin{pmatrix}3 \\ -10\end{pmatrix}$ and you're done. Note: This all assumes basis vectors of length 1. Things have to be scaled by the length of the basis vectors in the first coordinate system otherwise. Thanks from mosvas Last edited by romsek; December 10th, 2016 at 11:07 AM.
 December 10th, 2016, 03:07 PM #5 Newbie   Joined: Dec 2016 From: Natal - Brazil Posts: 10 Thanks: 0 I managed to find $\displaystyle Rp = \frac{10+3\sqrt{3}}{2}, \frac{3-10\sqrt{3}}{2}$ , but in the template is $\displaystyle \frac{-10+3\sqrt{3}}{2}, \frac{-3-10\sqrt{3}}{2}$ And how did you find it $\displaystyle v1 , v2$? Last edited by mosvas; December 10th, 2016 at 03:11 PM.

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